Directional Derivatives — Question 9

Question 9 – Solution

Strategy. Project the gradient onto the plane’s tangent subspace, then normalize the projection.

Step 1: Data $\nabla f(P)=\left\langle 2,4,4\right\rangle$. A unit normal to the plane is $$n=\frac 1{\sqrt 3}\left\langle 1,1,1\right\rangle.$$ The tangential projection is $$p=\nabla f-(\nabla f\cdot n)n =\left\langle 2,4,4\right\rangle-\frac{10}{3}\left\langle 1,1,1\right\rangle =\frac 13\left\langle-4,2,2\right\rangle.$$

Step 2: Direction and rate Since $\|p\|=2\sqrt 6/3$, $$\boxed{u_{\max}=\frac 1{\sqrt 6}\left\langle-2,1,1\right\rangle},\qquad \boxed{D_{\max,\mathrm{tangent}}f=\frac{2\sqrt 6}{3}}.$$

Step 3: Verify $u_{\max}\cdot\left\langle 1,1,1\right\rangle=0$, so it is tangent, and $\nabla f\cdot u_{\max}=2\sqrt 6/3$. If the projected gradient were zero, every tangent directional derivative would be zero and no unique maximizing tangent direction would exist.